#help-33
elfin berryBOT
mq
marsh citrusBOT
@noble harbor Has your question been resolved?
marsh citrusBOT
@noble harbor Has your question been resolved?
marsh citrusBOT
@noble harbor Has your question been resolved?
marsh citrusBOT
@noble harbor Has your question been resolved?
marsh citrusBOT
@noble harbor Has your question been resolved?
eternal hull
undone delta
!help
marsh citrusBOT
To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.
vital oracle
you cut out the important formula you were supposed to prove out of the image here
take another screenshot that includes the formula below the paragraph
green scaffold
marsh citrusBOT
green scaffold
idk why but my brain is not braining. I am stuck with this, does anyone see what I am doing wrong or does anyone have any suggestion on what to try instead?
devout mauve
in such cases its sometimes easier to prove something harder
for example, that the sum is > 13/24+1/n
green scaffold
for example, that the sum is > 13/24+1/n
thank you!
Is there any way to reasonably pick this extra term, or is it just intution and then trying if it holds?
weak surge
in such cases its sometimes easier to prove something harder
tighter*
weak surge
thank you!
Is there any way to reasonably pick this extra term, or is it just in...
because here you have three delicious terms which are less than 1/n FOR ALL N so to make the statement look clean and easier to visualise i assume @devout mauve chose that bound
marsh citrusBOT
green scaffold
.reopen
marsh citrusBOT
✅ Original question: #help-33 message
green scaffold
I was wondering, whether I am just doing something horribly wrong or if this new bound is actually not helpful, because it already fails for n=1?
just asking because it might have just been some examplary bound without much thought behind it, but also sometimes I mess up big time
marsh citrusBOT
Available help channel!
pulsar jolt
Can someone help me in GEMDAS/PEMDAS?
marsh citrusBOT
Can someone help me in GEMDAS/PEMDAS?
fringe iron
Can someone help me in GEMDAS/PEMDAS?
what do you mean? Like order of operations?
pulsar jolt
what do you mean? Like order of operations?
Yes exactly
fringe iron
what exactly do you not understand about it?
fringe iron
$10 + (2^{12-5} =32)*18 + 3$
elfin berryBOT
Katrro
fringe iron
like this?
stoic saddle
LIke this 10+[2(12-5)=32}18+3 with exponent 2
what's that equals sign doing in there..
can you show a picture of the problem as originally given
pulsar jolt
what's that equals sign doing in there..
idk
pulsar jolt
can you show a picture of the problem as originally given
sorry i cant iam on my loptop
the equal sign is + sorry
stoic saddle
loptop.
pulsar jolt
LIke this 10+[2(12-5)+32}18+3 with exponent 2
stoic saddle
$10+[2(12-5)+3^2]\cdot 18+3$
elfin berryBOT
Ann
stoic saddle
like this?
pulsar jolt
3 in the power of 2
stoic saddle
let's just put the multiplication back in explicitly: $$10+[2\cdot (12-5)+3^2]\cdot 18+3$$
elfin berryBOT
Ann
pulsar jolt
there is no 2 in 3
stoic saddle
can you confirm that the problem is written correctly or not
pulsar jolt
can you confirm that the problem is written correctly or not
it is send by a teacher
timber marlin
it is send by a teacher
why dont you start with easier ones first to understand the core of the concept
and then move on to harder ones
instead of directly jumping to a tricky q
timber marlin
send the exact question
fringe iron
I think the true question has been lost in translation
timber marlin
I think the true question has been lost in translation
he himself seems confused while sending the question
fringe iron
the real question is why does he have the undergraduate tag tho
pulsar jolt
10 + [2(12 -5) + 32} - 18 / 3^2
fringe iron
10 + [2(12 -5) + 32} - 18 / 3^2
write ^ for exponent
like 2^3 is 2 to the power of 3
pulsar jolt
18 divided by 3^2
pulsar jolt
parenthesis?
timber marlin
yep
so now only focus on this part first
and work your way out by solving the inner bracket first
once you do tell me what do you get as the result
timber marlin
what to do first parenthesis or brackets?
its the same idea!
so since theres a bracket inside, we solve that one first
because as i said, consider only that part so according to PEMDAS/BODMAS we solve the first inner bracket
pulsar jolt
10 + [2(7) + 32} - 18 / 3^2
timber marlin
yep
pulsar jolt
i get confuse 2 + 7?
timber marlin
i get confuse 2 + 7?
wheres 2+7?
pulsar jolt
2(7)
timber marlin
thats 2 x 7
pulsar jolt
10 + [14 + 32} - 18 / 3^2
pulsar jolt
10 + 46- 18 / 3^2
pulsar jolt
thats 3 x 3 right?
timber marlin
mhm
pulsar jolt
10 + [2(12 -5) + 32} - 18 / 9
timber marlin
- why did you write it in that form when we simplified it
and 2) why are the bracket notations different
pulsar jolt
oh sorry
pulsar jolt
10 + 46- 18 / 9
pulsar jolt
10 + 46- 2
timber marlin
there you go!
timber marlin
yup
timber marlin
how to close the problem?
.close
pulsar jolt
.close
balmy zinc
Can anyone please check if I did these questions correctly
marsh citrusBOT
Can anyone please check if I did these questions correctly
balmy zinc
hexed junco
Well the answers seem to be correct
balmy zinc
Well the answers seem to be correct
But did I do the work correctly or should I adjust anything?( also thank you )
tall iris
well like, maybe dont use the x for times/multiplication here, might confuse you
you can keep everything in brackets when multiplying, like (4)(x^3)(-13)
balmy zinc
you can keep everything in brackets when multiplying, like (4)(x^3)(-13)
Makes sense thank you for the tip I'll do that from now on
Nothing else ?
tall iris
looks good yeah
would just generally rec to use brackets when multiplying or $\cdot$, like $a \cdot b$
elfin berryBOT
aristos (kugelblitz)
balmy zinc
would just generally rec to use brackets when multiplying or $\cdot$, like $a \c...
Thank you so much
hexed junco
Can anyone please check if I did these questions correctly
Well the only minor problem i have here is you wrote (x-13)^4=(4C0)(x^4)(-13)^0
I would remove the (x-13)^4= at the 1st line and also at the last line write
(x-13)^4=(4C0)(x^4)(-13)^0 + (4C1)(x^3)(-13)^1+...
And then replace each terms with what you calculated
balmy zinc
Well the only minor problem i have here is you wrote (x-13)^4=(4C0)(x^4)(-13)^0
...
Ok thank you!
balmy zinc
Well the only minor problem i have here is you wrote (x-13)^4=(4C0)(x^4)(-13)^0
...
Wait actually I'm kinda confused sorry if you have time could you show me what that would look like
I can't think rn xd
marsh citrusBOT
@balmy zinc Has your question been resolved?
marsh citrusBOT
@balmy zinc Has your question been resolved?
marsh citrusBOT
@balmy zinc Has your question been resolved?
grave burrow
how do i approach this
hollow urchin
how do i approach this
Is this algebra 1?
grave burrow
Idk
hollow urchin
Ok, first solve for y
hollow urchin
Ok, remember that when 2 lines are parallel, they have the same slopes
grave burrow
Slope is -4
hollow urchin
Yep, now create any line with that slope
grave burrow
Downslopeing
hollow urchin
?
grave burrow
idk what u mean
if i were to make a line with that slope it would be downsloping is what i mean
hollow urchin
if i were to make a line with that slope it would be downsloping is what i mean
Yeah
Any line with that slope is parallel to the original line
grave burrow
yes
hollow urchin
So just make a random line with the same slope
marsh citrusBOT
marsh citrusBOT
Available help channel!
tired crescent
hi, what did i do wrong?
marsh citrusBOT
hi, what did i do wrong?
tired crescent
the rounding was confusing me a lot, i tried multiple different ways of rounding/displaying the answers and they still all showed incorrect
stoic saddle
when you said .03 did you mean three percent or did you mean tiny fraction of a percent
in letter a
tired crescent
uhhh idrk i just got that on the calculator, however i also tried 3 and that was still wrong
i can write it out to show how i did it if that would help
stoic saddle
yes, show how you did it. but also 
calculator-bashing...
stoic saddle
show how you did it
stoic saddle
im asking for the 3rd time SHOW how you did it
tired crescent
o. 12 is correct
omg
im so dumb
lemme try the next one again
ok i got a diff answer again. maybe my issue is that im not writing it out and doing it in my head
i got .13 for the second one
which would be 13 percent?
tired crescent
ok i redid them and got everything except 3
stoic saddle
so to put it as a percentage you will need to multiply by 100 to convert
for (c) pay attention to how much the interest is
it is 50 CENTS not 50 bucks
tired crescent
yes i used .50
tired crescent
is that right
okok
im redoing it rn one second
just to make sure i didnt miss any steps
i think im messing up rounding the last number
i keep getting the same one but im not sure how to enter that as an answer
or maybe i didnt do it right
sorry i forgot to reply to your message
stoic saddle
it says to round to the nearest tenth of a percent
which means that before multiplication by 100 that's 3 dp of rounding precision
0.171 in your case, or 17.1%
tired crescent
OOH I GET IT
I GOT IT!!
tysm for the help:DD
now onto the even harder version of this assignment LOL
i shall try it by myself and see what happens
.close
marsh citrusBOT
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gilded summit
because finite difference approximations are derived from a taylor series, are they only valid when the taylor series converges?
marsh citrusBOT
because finite difference approximations are derived from a taylor series, are t...
gilded summit
for example here
taylor series for f(x+d/2) and f(x-d/2) are used to find f'(x)
by using a small d such that the terms with a higher power of d can be ignored because in approximation
the document I am reading then goes on to mention the error by neglecting terms with power d^2 and greater
but they dont mention anything about whether the function is equal to its taylor seires, and just seem to assume
blazing pulsar
Usually a finite difference method only has a local smoothness condition. You don't need something as strong as what you are asking
gilded summit
Usually a finite difference method only has a local smoothness condition. You do...
why is this all you need?
gilded summit
but they dont mention anything about whether the function is equal to its taylor...
like i dont understand how they are able to just assume the function is equal to its taylor series in this derivation
blazing pulsar
I'm not sure what kind of response you are looking for here
You are usually looking to approximate a point in a neighbourhood with a linear or quadratic function, you don't really care about the Taylor series part
gilded summit
You are usually looking to approximate a point in a neighbourhood with a linear ...
isnt the taylor series how its derived so needs to be considered?
blazing pulsar
I mean, just consider any function that has some finite number of derivatives at a point
You can still do linear approximation on it
devout mauve
you can always do a finite taylor approximation
you dont have to do the full series
f(x+h)=f(x)+f'(x)h + R(h) is a taylor approx of first order
the rest term R(h) goes to 0 fast enough
gilded summit
You can still do linear approximation on it
ohh, because you are ignoring higher derivatives in the approximation, you dont need to care if the error goes to zero or not if you add all the terms?
fervent rampart
there are much less stringent conditions on when a function can be written as taylor polynomial + error term, and we only need to assume enough smoothness for whatever taylor polynomial we're using
devout mauve
the rest will go to zero fast enough
blazing pulsar
I just don't view it as a Taylor series at all, although Denascite's viewing of it being a finite Taylor approximation is the same.
devout mauve
well what else are you gonna call it
blazing pulsar
I just don't think you need to think of the language of Taylor series to approximate a function by a line or a quadratic etc
devout mauve
linear, ok
blazing pulsar
Yes, it turns out to be terms in a Taylor series, but I think in this case it's just confusing the issue
gilded summit
ohh, because you are ignoring higher derivatives in the approximation, you dont ...
this makes more sense now
blazing pulsar
Its a finite difference method, it's not like you are going to apply a 25th order adams method.
Anyway, its semantics and I don't really want to pursue that part of it further
gilded summit
thanks
gilded summit
this makes more sense now
it was a little confusing how they said it was equal to the function exactly without the remainder bit
!close
.close
marsh citrusBOT
marsh citrusBOT
Available help channel!
celest knoll
I dont think I graphed this right. Could someone help me
marsh citrusBOT
I dont think I graphed this right. Could someone help me
hazy lion
I dont think I graphed this right. Could someone help me
its $P(x) = (x-1)^2(x+2)^3$?
celest knoll
The x+2 is cubed
elfin berryBOT
jan Niku
celest knoll
Yes
hazy lion
this actually seems like it might be fine 
late geode
you have the key values,
looks a little off due to the vertical scale and placement of (2,64)
late geode
it's passable
celest knoll
Follow up question then
Why is one of the end behaviors in a negative Q if the leading coefficient is positive , that's what threw me off when I drew it
hazy lion
Why is one of the end behaviors in a negative Q if the leading coefficient is po...
can you track out what the leading term will be?
i mean you have $(x-1)(x-1)(x+2)(x+2)(x+2)$
elfin berryBOT
jan Niku
celest knoll
It would be x^5 but idk what the leading term would be
hazy lion
that is the leading term
x^5 + (Other lower degree stuff)
so x^5 dominates for really big (or really negative) x
celest knoll
Like the leading coefficient, what's that
hazy lion
the leading coefficient is 1
celest knoll
So if it's positive then why does my graph start in a negative Q
hazy lion
quadrant you mean?
celest knoll
Yea
hazy lion
well say x is really negative
now we have (negative) times (negative) times ...
maybe its better to write it
(negative) * (negative) * (negative) * (negative) * (negative)
you probably know that a negative times a negative is a positive?
so lets say [(negative) * (negative) ] * [(negative) * (negative) ] * (negative)
can you see what happens?
celest knoll
Yea
hazy lion
so for negative x, we get out a negative
celest knoll
Yea
hazy lion
at least, assuming that x^5 dominates
which will be true if x is really negative
'end behavior'
if that makes sense
celest knoll
Ok
Imma test
One sec
I tested at -3 and it was -16
So if I'm ever uncertain ig I can just do more test points
hazy lion
I think its better to rely on your intuition of end behavior
and test functions
but you could, yea
you run the risk that you do not grab a test point thats far enough out to be 'end behavior'
like, x^a for odd a is always going to look like this
,w plot x^7 and x^11
elfin berryBOT
marsh citrusBOT
marsh citrusBOT
Available help channel!
merry dust
how to prove this relation
marsh citrusBOT
how to prove this relation
merry dust
i started from the fact that :
$$ A \triangle B = (A \cap \neg B) \cup (\neg A \cap B)$$
elfin berryBOT
merry dust
and then distributed to get :
$$\left(E \cap (\neg A \cup \neg B) \right) \cap \left(E \cap (B \cup A) \right)$$
elfin berryBOT
marsh citrusBOT
marsh citrusBOT
Available help channel!
gritty zephyr
what is this question (#4) asking .. i dont understand
marsh citrusBOT
what is this question (#4) asking .. i dont understand
gritty zephyr
pls help, as a dad im supposed to know everything
main idol
How many combinations of red and blue backpacks add up to 10 backpacks
E.g. you can have 6 red and 4 blue
gritty zephyr
ahh, thats way easier to understand than the instructions thanks
marsh citrusBOT
Available help channel!
still temple
yo
still temple
i don’t understand how axioms 1 and 6 are violated
stoic saddle
what evidence have you to say either of these is specifically violated?
as in, who says that axiom 1 is violated specifically?
(which it's... not. that one holds)
dim lava
i don’t understand how axioms 1 and 6 are violated
well it also isnt clear what field the vector space is defined over
i guess the assumption is R
still temple
(which it's... not. that one *holds*)
the question
dim lava
in which case for c < 0, axiom 6 is violated
still temple
(which it's... not. that one *holds*)
i know it holds h thats why im confused
still temple
i guess the assumption is R
well yeah
dim lava
"Show that axiom 1 and 6 are violated"
still temple
show me where it says in writing 'Axiom 1 is violated'
it says “show where axioms 1 and 6 are violated”
stoic saddle
oh is that so
still temple
bruuhhhu
still temple
yes you can click on the picture
do they probably mean this
still temple
i have no clue atp
dim lava
i mean
fervent rampart
those aren't the axioms
still temple
not 1 and 6
how is 6 violated
still temple
right
dim lava
but the defn of the vector space says x >= 0
so -x is not in the set
nothing violates axiom 1 though
still temple
but the defn of the vector space says x >= 0
Yeah youre right
dim lava
nothing violates 1 though
for X = (x1, y1) and Y = (x2, y2)
X + Y (x1 + x2, y1 + y2)
Since x1 >= 0 and x2 >= 0, x1 + x2 >= 0 as well
etc
and then for y1 + y2 the sum of 2 reals is always real since R is complete yada yada yada
marsh citrusBOT
@still temple Has your question been resolved?
marsh citrusBOT
still temple
.reopen
marsh citrusBOT
✅ Original question: #help-33 message
still temple
i dont get what theyre asking
elfin cairn
You have R^2, but with a new scalar multiplication operation
ruby mulch
Do you know vector space axioms?
still temple
You have R^2, but with a new scalar multiplication operation
i don’t understand
ruby mulch
Let's focus on 6-10, can you reprove them in the new vector space? There should be at least one axiom that you can't prove.
still temple
Let's focus on 6-10, can you reprove them in the new vector space? There should ...
well i don’t understand the “new vector space” thing
theyre saying “scalar multiplication is redefined “
is that what it’s about?
ruby mulch
Then you use the new scalar multiplication instead of the older one. Simple.
still temple
Then you use the new scalar multiplication instead of the older one. Simple.
that means y is just 0?
or c is always 0 for y
ruby mulch
It means, well $c(x,y)=(cx, 0)$.
elfin berryBOT
Xwtek
still temple
Yeah but
Like why
is c set to 0 for y?
but then c is c so it must be always c right
if c isnt zero for x it shouldn’t be for y
still temple
but then y cant be 0 always because its any real number
ruby mulch
if c isnt zero for x it shouldn’t be for y
It's something that you're going to prove why or why not.
still temple
It's something that you're going to prove why or why not.
I thought i was just proving the invalidity of the space being a vector space
but im talking about the scalar redefinition
ruby mulch
You can try to define scalar multiplication by any function whatsoever. You just have to reprove that the vector space axiom is satisfied.
It's like saying f(x) = x^2 + 4x + 1, and then asking "why the constant is 1 instead of 4"
still temple
You can try to define scalar multiplication by any function whatsoever. You just...
No i get that but im asking what can that scalar be if y is any real number and c doesnt make x zero so it isnt 0
ruby mulch
The scalar is an argument to the operator. The one who uses it can set c to anything they wants
Like 2(x, y) = (2x, 0), 3(x, y) = (3x,0), etc.
still temple
Like 2(x, y) = (2x, 0), 3(x, y) = (3x,0), etc.
Yes but how did y become 0 here
ruby mulch
Definition
still temple
Definition
im confused
ruby mulch
The definition literally says c(x, y) = (cx,0)
still temple
The definition literally says c(x, y) = (cx,0)
but it needs to make sense though right
like what can x and y possibly be and what can c possibly be that would produce a non zero number for x and a zero number for y
ruby mulch
but it needs to make sense though right
It needs to follow vector space axiom... which is what you're trying to disprove here.
elfin cairn
like what can x and y possibly be and what can c possibly be that would produce ...
What you are missing is that this operation is not the same as the one you are used to
still temple
What you are missing is that this operation **is not the same as the one you are...
can you please explain what you mean by this?
ruby mulch
It literally says "scalar multiplication is redefined as ..."
fervent rampart
vector space = set + addition operation + scalar multiplication operation
if you change the addition or scalar multiplication operations you get a different vector space, even for the same set
ruby mulch
It means "forget old scalar multiplication and use this one"
still temple
But thats the thing i don’t understand
1- how “this new one” makes sense
And
2- how are we allowed to change definitions like that
ruby mulch
For example, $\mathbb{N}$ is either a trivial vector space, or you can redefine it by defining addition as multiplication and scalar multiplication as exponentiation.
elfin berryBOT
Xwtek
fervent rampart
you can define a vector space with any set, any addition operation, and any scalar multiplication operation (provided they follow the axioms). they don't have to be the "standard" operations
fervent rampart
For example, $\mathbb{N}$ is either a trivial vector space, or you can redefine ...
is N a vector space? over what field/operations?
ruby mulch
$\mathbb{N}$
elfin berryBOT
Xwtek
fervent rampart
N isn't a field?
elfin berryBOT
Xwtek
still temple
you can define a vector space with any set, any addition operation, and any scal...
but the scalar multiplication is how multiplication works.. like c * u = c(x,y) = (cx,cy) thats just how multiplication works right
fervent rampart
but the scalar multiplication is how multiplication works.. like c * u = c(x,y) ...
no, that's just a definition
you can define it a different way and get a different vector space
ruby mulch
It doesn't even have to be actual multiplication.
in the case of $\mathbb{Q}^+$, the scalar multiplication is exponentiation
elfin berryBOT
Xwtek
ruby mulch
wdym, $q^n$ is always rational for integer n
elfin berryBOT
Xwtek
ruby mulch
(Also, the vector addition here is actually multiplication)
fervent rampart
integers aren't a field
ruby mulch
I guess it's just a module.
But I think there should be an example where the scalar multiplication is not a regular multiplication.
still temple
you can define it a different way and get a different vector space
I don’t get this concept
I thought scalar multiplication and vector addition all followed a standard operation
and that same standard operation is what makes them a vector space
fervent rampart
no, they are part of the definition of the vector space along with the set
still temple
so there is a standard definition right
for scalar multiplication and vector addition
ruby mulch
And there is also "nonstandard" definition
still temple
But then the vector space is for the standard definition right
fervent rampart
But I think there should be an example where the scalar multiplication is not a ...
R^+ over R (or over Q if you want)
fervent rampart
But then the vector space is for the standard definition right
the vector space could be either. you just have to specify which operations you're talking about
unless otherwise specified you usually use the "standard" operations. but there's nothing special about them other than that they are common
still temple
R^+ over R (or over Q if you want)
oh theyre called “unusual vector space “ when you define the scalar multiplication or vector addition differently??
fervent rampart
unusual is not a technical term here
ruby mulch
It's just the title of the slide
fervent rampart
it is just describing the fact that it may seem strange to students
still temple
unless otherwise specified you usually use the "standard" operations. but there'...
because like usually if it isnt the standard way of using scalar multiplication or vector addition then that would violate the axiom
ruby mulch
But not always
fervent rampart
well certainly most "nonstandard" operations would violate the axioms. but some of them will follow the axioms and then they're perfectly valid as vector space operations
if you want to you can use something like $\oplus$ and $\odot$ to distinguish the nonstandard'' operations from the standard'' operations
elfin berryBOT
κλαουντ ☁ (cloud)
still temple
what do those symbols mean
fervent rampart
that doesn't change any of the math, just a notational difference
fervent rampart
what do those symbols mean
they mean addition and scalar multiplication, they're just circled to denote that they're "nonstandard" in some way
if using regular + and * makes you uncomfortable
fervent rampart
well if you were talking about R^2 being a vector space everyone would assume you mean using the standard operations, unless you specifically said otherwise
still temple
also if a vector space uses a non standard operation does that make it a different kind of vector space
fervent rampart
there's nothing mathematically different about vector spaces with nonstandard operations
only a difference in what's conventional
still temple
there's nothing mathematically different about vector spaces with nonstandard op...
wait how
still temple
R^+ over R (or over Q if you want)
clearly here they made k a power for example
fervent rampart
i mean that they're both vector spaces
vector spaces have the same fundamental properties no matter what the particular operations they use happen to be
still temple
vector spaces have the same fundamental properties no matter what the particular...
so if i had two sets, V and U and both had identical elements
But i chose to define the scalar multiplication and vec addition differently for both
They remain the same vector space?
If yes in what sense?
fervent rampart
no, they are different vector spaces
still temple
so whats the same thing that you were referring to
fervent rampart
i just mean there is nothing mathematical that says one vector space is "standard" and the other is "nonstandard"
ruby mulch
i mean that they're both vector spaces
.
fervent rampart
that's just based on convention
still temple
that's just based on convention
okay but in the problem, shouldn’t there be a mathematical logic that makes y always 0?
like I understand that if you wanna keep a valid vector space you can define scalar addition and scalar multiplication however you want unless it doesnt violate the axioms (hope this is correct)
but then i just wanna know what their definition means
Wait
i keep thinking of it as the standard way maybe I shouldn’t do that
fervent rampart
All the definition says is "to scalar multiply by a scalar c, multiply the first component by c and set the second component to 0"
you should verify which axioms are true if you follow that definition, and which are false
still temple
All the definition says is "to scalar multiply by a scalar c, multiply the first...
right but the setting to 0 part doesnt it need to follow some mathematical operation that makes sense?
like see how in c*(x,y) the components are literally being multiplied
= (cx,cy)
fervent rampart
you can view it as multiplying the second component by 0 if you want
still temple
which.. makes sense
still temple
you can view it as multiplying the second component by 0 if you want
but thats just so random
fervent rampart
right but the setting to 0 part doesnt it need to follow some mathematical opera...
whether an operation "makes sense" doesn't have anything to do with it. all it needs to do is produce a valid output for every input
still temple
like see how in c*(x,y) the components are literally being multiplied
or have the definitions always been irrelevant to how we use standard math?
Even in c*(x,y)?
fervent rampart
the vector space axioms don't care whether a function makes intuitive sense, they only care if it follows the axioms
marsh citrusBOT
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lilac stratus
Hi
lilac stratus
I have a math question in french 🇫🇷
stoic saddle
send it
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frank geyser
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blazing pulsar
I don't see how you are getting things like an i infront of partial u/partial y
You should probably write out the chain rule more carefully because it's hard to see what you are thinking
frank geyser
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blazing pulsar
You want f(z) = f(u,v) = u(x,y) + iv(x,y), not what you wrote
frank geyser
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blazing pulsar
This should have been explained in your book by the way, that u and v are real-valued functions with real variables x and y
frank geyser
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blazing pulsar
You are overcomplicating it by having this discussion instead of just doing the exercise
frank geyser
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blazing pulsar
what is partial f / partial u if not the thing you wrote right next to it.
frank geyser
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blazing pulsar
You still need the partial derivatives with respect to z
frank geyser
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blazing pulsar
The way I would do this is to write
$$\frac{\partial f}{\partial z} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial z} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial z}$$
then calculate $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ separately, then substitute it all in.
elfin berryBOT
JessicaK
blazing pulsar
The way you are doing this by trying to cram every part of the chain rule in a single line makes it very hard for me to follow
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faint grove
What is the length of PQ when □ABCD is folded as point A is on line BC and AE = 5,EB = 4,BP = 3?
marsh citrusBOT
What is the length of PQ when □ABCD is folded as point A is on line BC and AE = ...
cobalt sedge
whats P? whats Q? am I supposed to read it from your mind or smthn?
fervent tapir
where is Q?
cobalt sedge
post the full problem
faint grove
fluid tinsel
Gauss law
amber birch
triangle EBP should be similar to FRQ
cobalt sedge
Its helpful to track some angles in the figure and figure out the similarity
amber birch
triangle EBP should be similar to FRQ
yeah you can angle-chase it to see why
cobalt sedge
triangle EBP should be similar to FRQ
||PQC is better imo||
amber birch
||PQC is better imo||
fair enough
faint grove
triangle EBP should be similar to FRQ
There is no shared angle other than 90°?
amber birch
There is no shared angle other than 90°?
well, angle QCP is also 90
and importantly, what must angle EPQ be
cause where was angle EPQ originally before the folding?
faint grove
Angle DAE
amber birch
Angle DAE
yep!
so you can start with angle EPB = x
then chase to angle PQC and that should tell you something
marsh citrusBOT
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faint grove
Can't figure out what ratio are the area of EBP and FRQ
cobalt sedge
instead of FRQ, consider PCQ. That one is much easier to handle
marsh citrusBOT
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marsh citrusBOT
@faint grove Has your question been resolved?
calm harbor
I'd like my work checked
marsh citrusBOT
I'd like my work checked
calm harbor
We first set $x \in S_1 = (A\cap B)\cup(A\cap C)$ and $x \in S_2 = A\cap(B\cup C)$\
Let's analyze $S_1$ first.\
According to $S_1$, $x\in A$(otherwise it cannot be true whatsoever)\
x can be either in B or C\
Now for $S_2$\
Known fact: $x\in A$\
x also can be either in B or C.
elfin berryBOT
e4 e5 Qh5 Nc6 Bc4 Nf6 Qxf7#
calm harbor
Forgot to mention, either B or C has to contain x in both cases
I have a doubt, though. Shouldn't $S_1$ be equivalent to $S_2$?
elfin berryBOT
e4 e5 Qh5 Nc6 Bc4 Nf6 Qxf7#
indigo nest
I have a doubt, though. Shouldn't $S_1$ be equivalent to $S_2$?
They are yes
I imagine that's what this is building up to
calm harbor
thanks
Another quick question
So like range is "elements that have appear" and codomain is "elements that can possibly appear", right?
indigo nest
Yes
calm harbor
ah yes, that's clear
indigo nest
A bit categorically, codomain is the domain of f_inv
Where f: A → B and f_inv: B → P(A)
Every function is invertible under this definition of inverse
calm harbor
A bit categorically, codomain is the domain of f_inv
yep, learnt that b4
indigo nest
Nice nice
calm harbor
**e4 e5 Qh5 Nc6 Bc4 Nf6 Qxf7\#**
ye, is my thought process all right?
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calm harbor
You've done absorption laws right
yes, sry I have an emergency. Will be back later
Thanks for you time 
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indigo nest
yes, sry I have an emergency. Will be back later
No worries, feel free to tag me when you come back 🌺
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gloomy merlin
For known $\omega, A, B$, how do I go about finding out what $R$ and $\varphi$ are
[
A\6\cos{\omega t} + B\6\sin{\omega t} = R\6\cos{\omega t + \varphi}
]
marsh citrusBOT
For known $\omega, A, B$, how do I go about finding out what $R$ and $\varphi$ a...
elfin berryBOT
oak oxide
Trig
Or another method involves tranform Acos(wt) and Bsin(wt) to rotated vector around the origin
Which I forget its name
gloomy merlin
Rewrite in terms of complex exponentials, you mean?
stoic saddle
For known $\omega, A, B$, how do I go about finding out what $R$ and $\varphi$ a...
expand out R cos(ωt + φ)
you get $R \cos(\omega t) \cos(\varphi) - R \sin(\omega t) \sin(\varphi)$
elfin berryBOT
Ann
stoic saddle
therefore you should have $R\cos(\varphi) = A$ and $R \sin(\varphi) = -B$
elfin berryBOT
Ann
stoic saddle
this amounts to a translation from rectangular to cartesian coords
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white hull
how many quadruples of (A, B, C, D) exist such that there's a function f : Z -> Q that fulfills those conditions?
marsh citrusBOT
how many quadruples of (A, B, C, D) exist such that there's a function f : Z -> ...
white hull
i don't know where to even start
marsh citrusBOT
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white hull
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charred egret
Well you could start factoring out the quadratic
define S_1 = n^2-2n-1
and S_2 = n^2-4n+1
S_1 = (n-1)^2-2
s_2 = (n-2)^2-3
can look at two diff paramters give the same x
S_1(n) = S_1(m) = (n^2-2n-1) - (m^2-2m-1) = 0
simplify and n=m or m = 2-n
S_2 similiar idea
S_2(n) - S_2(m) = (n-m)(n+m-4)
stone pelican
@white hull , you can speak indonesian?
white hull
yeah
stone pelican
ok, jadi km udah tahu kah kalo 2022 = 47^2 -4*47 + 1
itu udh sangat berguna utk mengetahui C dan D sih sbg first step
white hull
oh jadi |47-c| + |47-D|=90
stone pelican
ya
stone pelican
how many quadruples of (A, B, C, D) exist such that there's a function f : Z -> ...
dan untuk poin ke1 dan ke 2, kalo n =1 maka dua2 nya menjadi f(-3)
jadi, f(-2) = (A+B)* (2+1/2) = |1-C|+ |1-D|
eh
wait i might be wrong
-2 sori
now the question menjadi jauh lebih simpel sih
i hope this gives you some good insight
white hull
jadi, f(-2) = (A+B)* (2+1/2) = |1-C|+ |1-D|
bukannya harusnya 2A+B/2?
white hull
i still don't really get how to continue though
stone pelican
lemme think
yg bikin susah tuh keknya |n-c| + |n-d| nya
jd mungkin kamu bisa memgabig menjadi 4 kasus
yaitu 47-c positif dan 47-d jg ,
47-c negatif
47-d negatif
dua2 negatif
jadi ga ada absolute valuenya lagi, aljabarnya lebih straightforward gitu
white hull
tapi kalau satunya negatif dan satunya positif fungsinya jadi konstanta kan?
stone pelican
lalu kalo tahu |47-c| + |47-d| = 90, nanti bisa mengetahui juga |1-c| + |1-d|
stone pelican
tapi kalau satunya negatif dan satunya positif fungsinya jadi konstanta kan?
eh
kek nya engga
yah kalo 47-c iitu negatif (sbg cth), |47-c| = c-47
oh iya iya
jadi c-d
yah its so easy now
2 kasus susah dan 2 kasus f(x) itu konstanta
dyk how to continue?
white hull
jadi kalau kasus 47-c negatif 47-d positif jadi C-D=90
stone pelican
jadi kalau kasus 47-c negatif 47-d positif jadi C-D=90
iya
white hull
hmm tapi kasus konstanta gak bisa kan karena kondisi pertama
stone pelican
ya i think
white hull
jadi antara dua duanya positif atau dua duanya negatif?
stone pelican
sorry i am kind of lagging
keknya yg 1 pos 1 neg itu bukan konstanta
i am so brainfoged rn
but i think so far the problem has been simplified quite a lot
so you should try to do it
white hull
ok i think i get the rest of the problem and i'm pretty sure the constant solution works too but just changing A and B
thanks for the help
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white hull
lol it's fine, the n^2-4n+1=2022 was something i would have missed
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rare owl
AAA i need help 😭 firstly im assuming I made the wrong choice for u because this didn’t simplify the integral one bit and js seems like a vicious loop atp if I try to integrate with parts again… can anyone tell me how to select u more wisely or if I made a mistake here 😞
oak oxide
You're on the right track really
Now u-sub the integral on the left just like what you did with the original
rare owl
I think if I use u=e^ax again it won’t get me anywhere in simplifying it though
oak oxide
I think if I use u=e^ax again it won’t get me anywhere in simplifying it though
Yes
But
rare owl
should i restart the question with u=sinbx instead
oak oxide
You will end up with the integral just the same as the original with different coefficient
I=..... + mI then you cam move I to the left side
I(1-m)=..... the right hand side won't have any integral
rare owl
I(1-m)=..... the right hand side won't have any integral
mm ok wait ill try it
fionna how do yk so much math 😭
i dunoo cuz I'm unemployed?
rare owl
sobs cuz wtaf
this is certainly smth
who wanna check my work for me 😊
i did this halfheartedly btw its prolly littered with errors
ill check it too 💔
odd orchid
should be ae^ax 2nd line
odd orchid
😐
sacred idol
isnt v = -cos(bx)/b?
marsh citrusBOT
sacred idol
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oak oxide
Yeah that's a legit crash out
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still temple
how can i express this in writing?
marsh citrusBOT
how can i express this in writing?
limber hearth
Well (fog)^-1 = g^-1 o f^-1 so yeah it commutes
fervent rampart
we would say that inverting distributes over multiplication composition if that equation were true in general
limber hearth
Well so yeah thats what cloud said
still temple
we would say that inverting distributes over ~~multiplication~~ composition if t...
ah i see
but is this multiplication? its composite
fervent rampart
typo, composition not multiplication
still temple
so i could phrase it as "the composition of two inverse functions is equivalent to the inverse of the composition in opposite order" ?
fervent rampart
that seems fine
we can actually note that this is true for any number of function compositions
still temple
so on top of that i could write note: this is true for any number of function compositions ? or just leave it as that
limber hearth
As you want ig
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sweet sparrow
how do i prove this without L'Hopital
marsh citrusBOT
how do i prove this without L'Hopital
turbid cargo
how do i prove this without L'Hopital
you know exp is growing a lot faster than x^beta
jagged relic
Take the log? Then it's basically x - log(x)
sweet sparrow
yea?
jagged relic
Then the derivative of that is positive increasing
sweet sparrow
yea i did that and got (ln (f(x)))' = a - b/x
then the derivative tends to a as x goes to infinity
something feels missing to me tho
jagged relic
a is positive, so there is a point past which that derivative is always positive
b is also positive, so that derivative is increasing
sweet sparrow
i think it's just me not confident about my answer
like there are functions that have positive derivative but does not tend to infinity
so it worries me
jagged relic
Because the positive derivative tends to zero
Here it tends to a which is positive, so intuitively the area under the curve gets infinitely large because the gap between the curve and the x-axis has a minimum width (past a certain point)
sweet sparrow
ye i think i can just write the derivative tends to a and hope that'd be an acceptable answer
blazing pulsar
The easiest way without Lhopital's is almost always going to be to Taylor expand if you ignore the fact that people then tend to run in circles arguing whether that is valid or not.
tardy citrus
How do I set up the double integral for parts a and b and how do I approach problems like these in general?
marsh citrusBOT
How do I set up the double integral for parts a and b and how do I approach prob...
tardy citrus
Here is my work but I know this is incorrect.
blazing pulsar
Why are you writing that 1/3 <= y_1
blazing pulsar
You want the probability that Y_1 is at most 1/2
there are numbers below 1/3 that you are elminating from contention
Like, what if y_2 is 1/100 for example, then y_1 can take on any value from 1/100 to 1/3 not captured in your writing
tardy citrus
okay that makes sense
blazing pulsar
You are integrating two things here, the square and the triangle, if you integrate dy2 first, then it isn't dy2 simple and you need to separate it into two integrals
if you do dy1 first, you can see it is dy1 simple
tardy citrus
what do you mean dy simple
blazing pulsar
I am surprised I cannot find a set of lecture notes on this online, but you should have had this in your Calculus course
A region is basically x or y simple if it is bounded from the top and the bottom by a single continuous function
I mean, technically the top is bounded by a piecewise continuous function, but you usually treat this case as two integrals, the one bounded by the line y = x and the one bounded by the line y = C
or in this notation, the one bounded by y_2 = y_1 and the one bounded by y_2 = 1/2
tardy citrus
so, y_2 is bounded at the top by y_1 and the bottom at 0
blazing pulsar
Yes, but only for the triangle part
tardy citrus
and since y_1 is bounded at the top at 1/2, the top bound of y_2 should be 1/2?
nah wait that actually doesnt sound right top bound is still y_1
blazing pulsar
Do you recognize that you are integrating the area of two shapes, a triangle and a square?
In one integral, the top is a diagonal line, and in the second, the top is a horizontal line
tardy citrus
so the triangle is 0 <= y_2 <= y_1 and 0 <= y_1 <=1/3
blazing pulsar
What you just wrote is a rectangle
tardy citrus
and the square is from 0 <= y_2 <= 1/3 and 1/3 <= y_1 <= 1/2
blazing pulsar
well, a square more precisely
tardy citrus
y_2 is bounded by y_2 = y_1 Is that correct
tardy citrus
What you just wrote is a rectangle
apologies
blazing pulsar
Yes
blazing pulsar
Alternatively, if you did dy_1 first, the top curve is just the line y_1 = 1/2 and the bottom line is y_1 = y_2
tardy citrus
Thank you so much
you are the best it all makes sense now
apologies for taking your time
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mossy widget
marsh citrusBOT
fervent rampart
,rcw
mossy widget
shit how do i rotate
elfin berryBOT
marsh citrusBOT
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mossy widget
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mossy widget
anyone????
mossy widget
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strange jungle
whats the topic called
marsh citrusBOT
@mossy widget Has your question been resolved?
hollow pine
is my way still valid?
I might be wrong but I think that $\nabla(\nabla\Phi(q))$ is not well defined. Usually what you are trying to write there is the Hessian of $\Phi$ that can be written as $\nabla\cdot\nabla^{T}$, where $\nabla$ is a column vector. However, if you take that into consideration, your answer should be the same. To be sure, if I had to write the way you did I would say that $\frac{d}{dt}(\nabla(\Phi(q)))=\dot{q}\cdot\frac{\partial}{\partial q}\nabla(\Phi(q))$
elfin berryBOT
KonoEmllikDa
mossy widget
I might be wrong but I think that $\nabla(\nabla\Phi(q))$ is not well defined. U...
yes that makes sense
i thought the same
as the gradient of the gradient is acc the laplace operator i think
hollow pine
The laplace operator for a scalar function is the divergence of the gradient
hollow pine
No problem what really matters is that you know how works
mossy widget
No problem what really matters is that you know how works
btw have u done lagrangian mechanics before?
hollow pine
But for a vector field is a little different, so you should be aware of it
hollow pine
btw have u done lagrangian mechanics before?
I wouldn't say I'm an expert, but I had a basic introduction to it in classical mechanics
If there is a problem you have you can send and if I can help I'll do my best
mossy widget
thx
its the final one
ive done the first thing, which is proving the time deriative equals 0
i can send my solution for that if u want
im now stuck on checking if this applies to the following lagrangian theyve given
mossy widget
If there is a problem you have you can send and if I can help I'll do my best
i started by differentiating L wrt q dot
hollow pine
Sure, I'm trying to do it, so can you show your work just to see if we are on the same track?
hollow pine
At least what you did seems right, we got the same result
hollow pine
for both
mossy widget
ok
idk how to go further now from the 2nd page
like im assuming u can simplify that derivative
hollow pine
If you use Euler-Lagrange equations on that Lagrangian you get that $q(\dot{q}\cdot\dot{q}+q\cdot\ddot{q})=0$, but after that I'm not finding a way to further simplifying it
elfin berryBOT
KonoEmllikDa
mossy widget
hmm
hollow pine
but you should try to apply EL to it and find it
mossy widget
but you should try to apply EL to it and find it
ok i will try thanks
mossy widget
but you should try to apply EL to it and find it
i will close it this channel as i need to log off
ty for ur help
i will lyk if i get an answer
hollow pine
Also the question seems to imply that you can't really show that it is equal to 0, the problem is showing that it's different to it, so if you can't do it, at least try to find the other conserved quantity
hollow pine
i will close it this channel as i need to log off
ty for ur help
i will lyk if...
Ok good luck with your study
mossy widget
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sturdy python
marsh citrusBOT
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limber hearth
how is it not c?
Ya un - devant
sturdy python
Ya un - devant
donc le b est negative?
limber hearth
Je sais pas ce que tu appelles b mais tu as que sqrt(x+5) est définie sur [-5,+inf[ et que donc en changeant le signe c'est sur ]-inf, -5]
sturdy python
Je sais pas ce que tu appelles b mais tu as que sqrt(x+5) est définie sur [-5,+i...
donc si il y avait pas de - c'est [-5,+inf[
limber hearth
Oui
sturdy python
le b c'est le chiffe avant le paranthese
ici c'est -1
donc le b est negative
tu comprend je veut dire quou? @limber hearth
limber hearth
Non mais l'important c'est que tu as compris pourquoi le domaine c'est d)
Est-ce le cas?
sturdy python
non parce que je comprend pas le truc du signe
le b ici
il y a un chiffre ici
-1
limber hearth
Et bien tu résoud une inéquation
Faut que la quantité sous la racine sois positif
Donc -(x+5) >= 0
Puis de ça tu obtiens ton intervalle
sturdy python
si c'etait -(-x +5) on fait quoi
limber hearth
La même chose
sturdy python
donc la courbe va vers la gauche?
limber hearth
Mais avec -(-x+5)
limber hearth
donc la courbe va vers la gauche?
Comme ça : \ ?
limber hearth
Alors oui
sturdy python
ok c sa le truc que je comprend pas
quand jai appris les racine carree
quand le a est positive et le b est positive le courbe va en haut a droite
regade
limber hearth
La racine carrée ne change en rien la fonction qui est dedans
sturdy python
le a est positif
mais le b est negative
b = -1
pourquoi sa va en bas
c'est en haut a gauche
limber hearth
donc a*sqrt(bx + c) ?
limber hearth
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marsh citrusBOT
✅ Original question: #help-33 message
sturdy python
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limber hearth
Ok admettons
sturdy python
est ce que le a est positif ou negative
limber hearth
4sqrt(-x-5) c'est la même chose que sqrt(-2x-10) oui ?
sturdy python
oui
limber hearth
Bon et la racine est strictement croissante donc elle n'influence pas la "forme" de la courbe si tu veux
sturdy python
comment on sais le elle est strictement croissant
limber hearth
Donc enfaîte sqrt(-2x-10) va "aller du même côté" que -2x-10
sturdy python
Bon et la racine est strictement croissante donc elle n'influence pas la "forme"...
comment on sais que elle est strictement croissante?
sturdy python
k yo donc si c'est f(x) = 4sqrt(x + 5) − 1?
comment est ce que sa va vers le gauche
quand c'est -
ta dit le direction change
donc sa va comme sa \
comment est ce que meme sans le signe - sa va encore /
sturdy python
ok mais on peut pas avoir un racine negatif
sturdy python
Parce que -x-5 va "à gauche"
ok mais c'est (x+5)
pas -x -5
donc sa va quelle direction
a droite? \
limber hearth
Et bien dans ce cas ça va du côté de x+5
sturdy python
On dit que la fonction est croissante
comment on sais
limber hearth
comment on sais
Si x <= y => f(x) <= f(y) alors la fonction est croissante
De manière plus générale si f est de la forme ax+b alors f est croissante si et seulement si a est positif
Ici on a x+5
Donc a = 1 >= 0
Donc croissante
Et même strictement croissante si a > 0 (vrai pour x+5 donc)
limber hearth
Pour -x-5 c'est strict décroissant puisque a = -1 <= 0
sturdy python
Pour -x-5 c'est strict décroissant puisque a = -1 <= 0
mais c'est sa l'equation
c'est -(x+5)